Math, asked by bostonpolglasemarsh, 27 days ago

Alice’s age is currently the same as the sum of Bella’s and Charles’ ages.
Ten years ago it was the case that Alice was twice as old as Bella was then.
In how many years time will it be the case that Alice will be twice as old as Charles?

Answers

Answered by RvChaudharY50
0

Solution :-

Let Alice present age is A years , Bella present age is B years and Charle present age is C years .

so,

→ A = (B + C) ----- Eqn.(1)

and, 10 years ago,

→ (A - 10) = 2(B - 10)

→ A - 10 = 2B - 20

→ 2B = A + 10

→ B = (A + 10)/2 ------ Eqn.(2)

putting value of Eqn.(2) in Eqn.(1) ,

→ A = (A + 10)/2 + C

→ A = (A + 10 + 2C)/2

→ 2A - A = 2C + 10

→ A = 2C + 10

→ A = 2(C + 5) ------ Eqn.(3)

now, let after x years A will be twice as old as charles .

So,

→ A + x = 2(C + x)

putting value of Eqn.(3) in LHS,

→ 2(C + 5) + x = 2(C + x)

→ 2C + 10 + x = 2C + 2x

→ 2x - x = 10

→ x = 10 years (Ans.)

Hence, after 10 years Alice will be twice as old as Charles .

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